Derivative at a Point — Meaning, Notation and Examples
Students often learn the differentiation rules before they learn what a derivative actually is. That works until a question asks for the derivative "from first principles", or for the equation of a tangent, or whether a point is a maximum — and then the rules alone are not enough. This article explains what f′(a) means, where the definition comes from, how to read the notation, and shows four worked examples with every step.
The definition
The derivative of a function f at the point x = a is defined as a limit:
An equivalent form, useful in some proofs, is:
| Symbol | What it means |
|---|---|
| f(a) | The value of the function at x = a — a point on the curve |
| h | A small step away from a; it is made to approach 0, never set equal to 0 |
| [f(a+h) − f(a)] / h | The slope of the straight line (a secant) joining two points on the curve |
| f′(a) | A single number — the slope of the tangent at x = a |
| f′(x) | A function — the slope at every x, also written dy/dx |
| f″(a) | The derivative of f′ evaluated at a — how fast the slope itself is changing |
Notice why we cannot just put h = 0: the fraction becomes 0/0, which is meaningless. The limit is a way of asking what the ratio settles down to as h shrinks, without ever letting it reach zero.
The geometric meaning
Take two points on the curve: (a, f(a)) and (a + h, f(a + h)). The straight line through them has slope [f(a+h) − f(a)] / h — that is just "rise over run". Now slide the second point closer and closer to the first. The secant line pivots, and in the limit it becomes the tangent — the line that touches the curve at that one point and matches its direction there. So:
Tangent line: y − f(a) = f′(a) (x − a)
The physical reading is the same idea in different clothes. If s(t) is displacement, s′(t) is velocity and s″(t) is acceleration. If C(x) is cost, C′(x) is marginal cost. A derivative is always "how fast one quantity changes when another changes".
Standard derivatives you should know by heart
| f(x) | f′(x) |
|---|---|
| c (a constant) | 0 |
| xⁿ | n xn−1 |
| sin x | cos x |
| cos x | −sin x |
| eˣ | eˣ |
| ln x (x > 0) | 1 / x |
Worked example 1 — from first principles
Question: Find f′(3) for f(x) = x², using the definition.
f(3 + h) = (3 + h)² = 9 + 6h + h²
f(3) = 9
f′(3) = limh→0 [ (9 + 6h + h²) − 9 ] / h
= limh→0 (6h + h²) / h
= limh→0 (6 + h) [cancel h, valid because h ≠ 0]
= 6
Check with the rule: f′(x) = 2x, so f′(3) = 2(3) = 6 ✔
Tangent line at x = 3: f(3) = 9, so
y − 9 = 6(x − 3) → y = 6x − 18 + 9 → y = 6x − 9
Worked example 2 — a cubic, with the second derivative
Question: For f(x) = x³ − 5x + 2, find f′(2), f″(2), and the tangent at x = 2.
Differentiate term by term:
f′(x) = 3x² − 5
f″(x) = 6x
Evaluate at x = 2:
f′(2) = 3(2)² − 5 = 3(4) − 5 = 12 − 5 = 7
f″(2) = 6(2) = 12
The point on the curve: f(2) = (2)³ − 5(2) + 2 = 8 − 10 + 2 = 0. So the curve passes through (2, 0).
Tangent: y − 0 = 7(x − 2) → y = 7x − 14
Reading the answers: f′(2) = 7 is positive, so the function is increasing at x = 2. f″(2) = 12 is positive, so the curve is concave up (bending upward) there.
Worked example 3 — chain rule at a point
Question: Find f′(1) for f(x) = (2x + 1)⁴.
The outer function is (something)⁴ and the inner function is 2x + 1, so differentiate the outside and multiply by the derivative of the inside:
f′(x) = 4(2x + 1)³ × 2 = 8(2x + 1)³
At x = 1: 2x + 1 = 2(1) + 1 = 3, and 3³ = 27, so
f′(1) = 8 × 27 = 216
Why the ×2 matters: leaving it out gives 4 × 27 = 108, exactly half the correct answer. Forgetting the inner derivative is the single most common calculus error in board answer sheets.
Worked example 4 — the second derivative test
Question: Find and classify the turning points of f(x) = x³ − 3x.
Step 1 — set the first derivative to zero:
f′(x) = 3x² − 3 = 0 → 3(x² − 1) = 0 → x = 1 or x = −1
Step 2 — test with the second derivative: f″(x) = 6x
f″(1) = 6(1) = 6 > 0 → concave up → local minimum at x = 1
f″(−1) = 6(−1) = −6 < 0 → concave down → local maximum at x = −1
Step 3 — find the values:
f(1) = (1)³ − 3(1) = 1 − 3 = −2 (local minimum value)
f(−1) = (−1)³ − 3(−1) = −1 + 3 = 2 (local maximum value)
So the curve has a maximum at (−1, 2) and a minimum at (1, −2). Note the maximum value here is smaller than the minimum value on the other branch — "local" really does mean local.
When the derivative does not exist
A function must be continuous at a to be differentiable there, but continuity alone is not enough. Take f(x) = |x| at x = 0. Approaching from the right, the slope is +1; from the left it is −1. The two one-sided limits disagree, so f′(0) does not exist — the graph has a sharp corner, and no single tangent can be drawn. The same happens at a vertical tangent, such as f(x) = x1/3 at x = 0.
Common mistakes that cost marks
- Substituting the value before differentiating. If you put x = 2 into f(x) = x³ − 5x + 2 first, you get the number 0, whose derivative is 0. Always differentiate the general function, then substitute.
- Confusing f′(a) with f(a). f(a) is a height on the graph, f′(a) is a slope. A tangent question needs both.
- Forgetting the inner derivative in the chain rule — see example 3.
- Assuming f′(a) = 0 always means a maximum or minimum. For f(x) = x³ at x = 0, f′(0) = 0 and f″(0) = 0, but it is a point of inflection, not a turning point. When f″(a) = 0 the test is inconclusive and you must check the sign of f′ on both sides.
- Cancelling h without stating h ≠ 0. Inside the limit h is never actually zero, which is exactly what makes the cancellation legal. Examiners look for this line in first-principles answers.
- Writing the tangent as y = f′(a)x + f(a). The correct form is the point-slope form y − f(a) = f′(a)(x − a).
Where this appears in exams
| Exam | Typical use |
|---|---|
| CBSE/ICSE Class 11 | Limits and derivatives — first-principles derivations carry full marks of their own |
| CBSE/ICSE Class 12 | Application of derivatives — tangents, normals, increasing/decreasing, maxima and minima |
| JEE Main & Advanced | Differentiability at a point, chain rule chains, optimisation problems |
| Physics (any level) | Velocity as ds/dt, acceleration as d²s/dt², rate-of-change problems |
| Physical chemistry | Rate laws, slopes of concentration–time and Arrhenius plots |
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